What is f(x) = int e^x-x dxf(x)=∫ex−xdx if f(-1) = 1 f(−1)=1?

1 Answer
Bill K.
Feb 24, 2016

f(x)=e^(x)-x^2/2+3/2-e^{-1}f(x)=ex−x22+32−e−1

Explanation:

The general antiderivative is

\int (e^{x}-x)\ dx=e^{x}-x^{2}/2+C

Therefore, f(x)=e^{x}-x^{2}/2+C for some constant C. Since f(-1)=1, we get 1=e^{-1}-(-1)^{2}/2+C=e^{-1}-1/2+C, which implies that C=3/2-e^{-1}.

Hence, f(x)=e^(x)-x^2/2+3/2-e^{-1}.

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